A homogoneous chain of length L lies on a table. The coefficient of friction between the chain and the table is `mu`. The maximum length which can hang over the table in equilibrium is

Options

A. `((mu)/(mu+1))L`
B. `((1-mu)/(mu))L`
C. `((1-mu)/(1+mu))L`
D. `((2mu)/(2mu+1))L`

Answer

To solve the problem of finding the maximum length of a homogeneous chain that can hang over a table in equilibrium, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Variables**:
- Let the total length of the chain be \( L \).
- Let the length of the chain hanging over the table be \( L' \).
- The length of the chain on the table will then be \( L - L' \).
- The coefficient of friction between the chain and the table is \( \mu \).

2. **Understand the Forces Acting on the Chain**:
- The weight of the hanging part of the chain (mass \( m' \)) is given by:
\[ m' = \frac{m}{L} \cdot L' \]
where \( m \) is the total mass of the chain.
- The weight acting downwards due to the hanging part is:
\[ F_{\text{weight}} = m' g = \left(\frac{m}{L} \cdot L'\right) g \]

3. **Frictional Force**:
- The frictional force \( F_{\text{friction}} \) acting on the part of the chain on the table is given by:
\[ F_{\text{friction}} = \mu N \]
where \( N \) is the normal force.
- The normal force \( N \) is equal to the weight of the part of the chain on the table:
\[ N = \left(\frac{m}{L} \cdot (L - L')\right) g \]
- Therefore, the frictional force becomes:
\[ F_{\text{friction}} = \mu \left(\frac{m}{L} \cdot (L - L')\right) g \]

4. **Equilibrium Condition**:
- For the chain to be in equilibrium, the frictional force must balance the weight of the hanging part:
\[ F_{\text{friction}} = F_{\text{weight}} \]
- Substituting the expressions for the forces:
\[ \mu \left(\frac{m}{L} \cdot (L - L')\right) g = \left(\frac{m}{L} \cdot L'\right) g \]

5. **Cancel Common Terms**:
- Cancel \( g \) and \( \frac{m}{L} \) from both sides:
\[ \mu (L - L') = L' \]

6. **Rearranging the Equation**:
- Rearranging the above equation gives:
\[ \mu L - \mu L' = L' \]
- Combine like terms:
\[ \mu L = L' + \mu L' \]
\[ \mu L = L' (1 + \mu) \]

7. **Solving for \( L' \)**:
- Finally, solve for \( L' \):
\[ L' = \frac{\mu L}{1 + \mu} \]

### Conclusion:
The maximum length of the chain that can hang over the table in equilibrium is:
\[ L'_{\text{max}} = \frac{\mu L}{1 + \mu} \]